Question

how do i do this problem

(3/4)^2 x (2/3)^3

i got 1/4 but in the book it says 1/6

Answer 1

opps, i got 1/6 not 1/4

Answer 2

opps i meant to write 1/8 omg

Answer 3

\(\bf \left(\cfrac{3}{4}\right)^2\times \left(\cfrac{2}{3}\right)^3\implies
\cfrac{3^2}{4^2}\times \cfrac{2^3}{3^3}\implies \cfrac{\cancel{ 3^2 }}{(2^2)^2}\times \cfrac{2^3}{\cancel{ 3^2 }3^1}\implies \cfrac{1\times 2^3}{2^4\times 3^1}
\\ \quad \\
\cfrac{1\times \cancel{ 2^3 }}{\cancel{ 2^3 }2^1\times 3^1}\implies ?\)

Answer 4

@jdoe0001- thank you for your answer but i find it way to confusing...can i write you my work and you can tell me where i went wrong at..please

Answer 5

you do know that \(\bf 3^3\to 3^2\cdot 3^1\to 3^{2+1}\to 3^3\) right? is all I did... I factor it, and thus the factors cancel out

Answer 6

\[\left( \frac{ 3}{ 4 } \right)^{2} \times \left( \frac{ 2 }{ 3 } \right)^{3} = \frac{ 9 }{ 16 } \times \frac{ 6 }{ 7 } = \frac{ 63 }{ 432 } = \frac{ 7 }{8}\]

Answer 7

:0 im lost :(

Answer 8

ok

Answer 9

lemme use units then

Answer 10

\(\bf \left(\cfrac{3}{4}\right)^2\times \left(\cfrac{2}{3}\right)^3\implies
\cfrac{3^2}{4^2}\times \cfrac{2^3}{3^3}\implies \cfrac{3^2\times 2^3}{4^2\times 3^3}
\\ \quad \\
\cfrac{\cancel{ 3\times 3 }\times 2\times 2\times 2}{4\times 4\times \cancel{ 3\times 3 }\times 3}
\implies \cfrac{ 2\times 2\times 2}{4\times 4\times 3}\implies \cfrac{\cancel{2\times 2\times 2 }}{\cancel{2\times 2\times 2 }\times 2\times 3}\implies ?\)